The homotopy groups of the étale topos of a scheme were defined by Artin and Mazur. Are these known for Spec Z? Certainly π1 is trivial because Spec Z has no unramified étale covers, but what is known about the higher homotopy groups?

3 Answers
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$Spec(\mathbb{Z})$ should only be considered as $S^3$, if you "compactify" that is add the point at the real place. This is demonstrated by taking cohomology with compcat support.
The étalé homotopy type of $Spec(\mathbb{Z})$ is however contractible (indeed what do you get by removing a point form a sphere?) to see this (all results apper in Milne's Arithmetic Dualities Book)

is the first line in your answer referring to some deleted comment? The way it's phrased it sounds like you are responding to some remark but I can't seem to find anything relevant. Anyway, I would like to understand why "compactified" Spec Z should be the 3-sphere. thanks!
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bananastackNov 4 '14 at 4:14

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@user125763, My remark is referring to the comment by Ilya Nikokoshev. In any case the dulaity above, presents
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Tomer SchlankNov 4 '14 at 6:58

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The duality above, presents $X$ as "a non-orinted 3-manifold with boundry" by analogy with poincare duality. One can also consider the analogy with curves over finite field which in étale eyes are a 2-dim surface fiberation over a circle (but only if taking all points!). This analogy is very power full, for e.g you can think about primes as circles embedded in the 3-sphere (knots) and then the Legendre symbol is related to the linking number. For more about this philosophy, read in "knots and primes" by Morishita
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Tomer SchlankNov 4 '14 at 7:12

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@Joël, the hurwitz Theorem and the profiniteness theorem for \'etale homotopy" are in the Artin-Mazur's paper. The universal coeff theorem, is not explicitly stated (i think) but it follows from the regular universal coeff theorem since the cohomology is cohomology of a pro-space
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Tomer SchlankNov 4 '14 at 21:00

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The best source on those that I am aware of is in Quick's work which builds up homotopy theory of simplicial profinite sets. Specifically the papers "Profinite Homotopy Theory" and "Continuous group actions on profinite spaces", note that there were minor errors in the former corrected in the latter. But in that formalism one does get homology, cohomology, higher homotopy groups, a Hurewicz map. He doesn't prove a Kunneth formula but if I recall correctly that's just a spectral sequence argument and should follow formally from the other pieces. There are a few more subtleties too, of course.
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Joe BNov 4 '14 at 21:00

Since nobody else is answering this question I will give a few vague thoughts. Caveat:
this is a completely speculative answer for various reasons, not least of which because I don't know Artin-Mazur's definition of pi_i.

Since X = Spec(Z) is "simply connected", one can pretend that the Hurewicz theorem applies.

I believe that H^i(X,Z/nZ) is trivial for all i, as a consequence of class field theory and the fact that Z has neither many units nor n-th roots of unity. (I'm not completely sure about i = 3 and n = 2 here.)
One can then squint and imagine that the higher homotopy groups of X are trivial. This seems a little dodgy. Another direction one could go is to note that the groups H^i(X,G_m) vanish unless i = 3, and H^3(X,G_m) = Q/Z. From this (and other) facts it has been argued that X is analogous to the 3-sphere.

For what it is worth, both computations suggest that pi_2(X) is trivial. If one wanted to turn this comment into mathematics, one should try to define an algebraic Hurewicz map.

Indeed H^i(X) is supposed to point out that X looks like a 3-sphere and it's not like homotopy groups of S^3 are easy.
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Ilya NikokoshevNov 3 '09 at 9:03

My memory is vague, but does H^3(X,G_m) = Q/Z imply that there are nontrivial elements in H^3(X,Z/2) arising from the kernel of the squaring map G_m -> G_m, since Z/2 and the 2nd roots of unity represent the same etale sheaf on this site?
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Tyler LawsonNov 3 '09 at 12:23

The word "represent" may not have been the best to use. No, it's not etale, but it's a sheaf on the etale site as the kernel of the map G_m -> G_m. (The cokernel, if I remember correctly, is isomorphic the direct image of the additive group on the etale site of Z/2.)
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Tyler LawsonNov 3 '09 at 15:02

Oh, I see what you mean. Thanks for clarifying!
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moonfaceNov 3 '09 at 17:20

If etale pi_1 classifies obstructions to trivializing finite flat unramified Z-algebras, it would be nice if the whole etale homotopy type classified obstructions to trivializing simplicial commutative Z-algebras that were finite, flat, and unramified in a homotopy sense. I think all of these notions make sense: "finite" means that the homotopy groups vanish in high degrees and are finitely generated, "flat" means that these homotopy groups have no torsion, and "unramified" means that the cotangent complex is zero. Is that right?

Presumably algebraic topologists have thought about the sphere spectrum version of this question. Are there any connective E-infinity ring spectra that are finite, flat, and unramified over the sphere?

After Tyler's comments, I see that this is a bad analogy. The dictionary between etale locally constant sheaves of sets and finite flat unramified algebras (which in one direction takes an algebra and associates the sheaf of sections of its spectrum over Spec Z) just doesn't extend to a dictionary between homotopy-style locally constant sheaves and homotopy-style finite flat unramified algebras.

With regards to your second question: This appears in John Rognes' paper on Galois theory of structured ring spectra, at least in part. There are none if you ask that the generators all live in pi_0. If you allow new generators in positive degrees, there are square-zero extensions and their ilk, and more following those that are harder to classify. I am not sure about your first question.
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Tyler LawsonNov 3 '09 at 21:52

Are you really saying that there are square-zero extensions of the sphere spectrum that are unramified? This isn't possible with plain rings.
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David TreumannNov 3 '09 at 22:55

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I guess I was thinking by analogy with your "homotopy vanishing in high degrees" examples above. No, those can't ever have trivial cotangent complex, and thickenings of a commutative Z-algebra to something with positive homotopy groups usually can't either; e.g. if R -> S is an isomorphism on pi_0 then the first nonvanishing relative homotopy group coincides with the first nonzero homotopy group of the cotangent complex.
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Tyler LawsonNov 4 '09 at 0:35